'''simple docstring'''




from itertools import permutations






def __lowerCAmelCase					(				UpperCamelCase__							)      ->       bool:
    if num[3] % 2 != 0:
        return False

    if (num[2] + num[3] + num[4]) % 3 != 0:
        return False

    if num[5] % 5 != 0:
        return False

    __lowerCamelCase            =    [7, 11, 13, 17]
    for i, test in enumerate(UpperCamelCase__							):
        if (num[i + 4] * 1_00 + num[i + 5] * 10 + num[i + 6]) % test != 0:
            return False
    return True






def __lowerCAmelCase					(				UpperCamelCase__ = 10							)      ->       int:
    return sum(
        int(''''''.join(map(UpperCamelCase__							,    UpperCamelCase__							)							)							)
        for num in permutations(range(UpperCamelCase__							)							)
        if is_substring_divisible(UpperCamelCase__							)							)


if __name__ == "__main__":
  print(f'{solution() = }')

'''simple docstring'''




def __lowerCAmelCase					(				UpperCamelCase__							,    UpperCamelCase__							)      ->       float:
    return base * power(UpperCamelCase__							,    (exponent - 1)							) if exponent else 1


if __name__ == "__main__":
  print("Raise base to the power of exponent using recursion...")
  __UpperCAmelCase					=int(input("Enter the base: ").strip())
  __UpperCAmelCase					=int(input("Enter the exponent: ").strip())
  __UpperCAmelCase					=power(base, abs(exponent))
  if exponent < 0:  # power() does not properly deal w/ negative exponents
    __UpperCAmelCase					=1 / result
  print(f'{base} to the power of {exponent} is {result}')
